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id="content-inner"><div id="post"><article class="post-content" id="article-container"><h1 id="1"><a href="#1" class="headerlink" title="1"></a>1</h1>

用反证法. 假设 $a+b$ 是有理数, 那么有 $a+b = \frac{p}{q}$ , 令 $a=\frac{p'}{q'}$ , 那么有
$$
\begin{aligned}
\frac{p'}{q'} + b &= \frac{p}{q}\\
b&=\frac{pq'-p'q}{qq'}
\end{aligned}
$$
即 $b$ 为有理数, 与题目矛盾, 因此 $a+b$ 是无理数. $a-b$ 是同理可得.

若 $ab$ 是有理数, 那么有 $ab = \frac{p_0}{q_0}$ , 因此 $b=\frac{p_0}{aq_0}$ , 即 $b$ 为有理数, 矛盾, 因此 $ab$ 是无理数. $a/b$ 同理可得.



<h1 id="2"><a href="#2" class="headerlink" title="2"></a>2</h1>

这里只考虑大于零的情况 (小于等于零的情况类似, 因此为了方便不做讨论) .  假设有两个有理数 $a=\frac{p}{q}>b=\frac{p'}{q'}>0$ . 通分得 $\frac{pq'}{qq'}>\frac{p'q}{qq'}$ , 显然有 $\frac{pq'-1}{qq'}\geqslant\frac{p'q}{qq'}$ , 等价于 $\frac{kpq'-k}{qq'k}\geqslant\frac{p'qk}{qq'k}$ , 其中 $k\in\mathbb Z^*$因此 $a,b$ 两个有理数之间至少有 $\frac{kpq'}{qq'k}, \frac{kpq'-1}{qq'k},\dots,\frac{kpq'-k+1}{qq'k}$ , $k$ 个有理数, 有由于 $k$ 可以取任意大, 因此 $a,b$ 间有无限多个有理数. 而 $\frac{pq'}{qq'}>\frac{\sqrt{(pq')^2-1}}{qq'}>\frac{p'q}{qq'}$ , 由该节知  $\sqrt{(pq')^2-1}$ 是无理数, 由练习题 1 知 $\frac{\sqrt{(pq')^2-1}}{qq'}$ 是无理数, 因此两个有理数间至少有一个无理数. 由于两个有理数间有无限多个有理数, 因此两个有理数间也就有无限多个无理数.



<h1 id="3"><a href="#3" class="headerlink" title="3"></a>3</h1>

用类似书里的方法, 假设 $\sqrt[3]{2} = \frac{p}{q}$ , $p$ 与 $q$ 不存在公因子, 那么 $\sqrt[3]{2}q=p$ , 等价于 $2q^3=p^3$ , 显然 $p$ 是偶数, 令 $2p'=p$ , 那么 $q^3=8p^3$ , 因此 $q$ 也是偶数, 产生矛盾, 于是 $\sqrt[3]{2}$ 是无理数.



<h1 id="4"><a href="#4" class="headerlink" title="4"></a>4</h1>

平方得 $5+2\sqrt{6}$ , 显然是无理数, 假设 $\sqrt 2+\sqrt 3=\frac{p}{q}$ , 那么 $\frac{p^2}{q^2}$ 就是无理数了, 这显然不正确, 于是 $\sqrt 2+\sqrt 3$ 是无理数.



<h1 id="5"><a href="#5" class="headerlink" title="5"></a>5</h1>

一开始看这道题我就估计这与 $\sqrt{2}$ 是无理数有关, 果然猜对了.

显然 $(0,0)$ 是该圆上的一个有理点. 若 $\frac{p}{q},\frac{p'}{q'}\not=0, (\frac{p}{q},\frac{p'}{q'})$ 是该圆的一个有理点, 那么有 ${\left(\frac{p}{q}\right)}^2-{\left(\frac{p'}{q'}\right)}^2 -2\sqrt{2}\frac{p}{q}=0$ , 即 $2\sqrt 2=\frac{{\left(\frac{p}{q}\right)}^2-{\left(\frac{p'}{q'}\right)}^2}{\frac{p}{q}}$ 显然是有理数 (懒得化简) , 与 $\sqrt{2}$ 是无理数矛盾, 因此不存在 $(0,0)$ 之外的有理点, 即该圆只有唯一有理点.



<h1 id="6"><a href="#6" class="headerlink" title="6"></a>6</h1><p>小学的竖式除法派上用场啦! 同样只考虑正数. 由于有尽小数可以看成是特殊无尽循环小数, 因此可以合并处理. 考虑 $\frac{p}{q}$ , 我们用竖式除法来做, 第 $n$ 次除得到的数 (第一个非零数开始算起) 称为 $a_n$ , 余下的数称为 $b_n$ (已经进位). 比如说 22/7 , 那么 $a_1=3, b_1=10, a_2=1, b_2=30$ .</p>

$$
\begin{aligned}
22/7&=3\cdots\cdots1 \quad(进位变为10)\\
10/7&=1\cdots\cdots3 \quad(进位变为30)
\end{aligned}
$$


<p>显然对于固定的 $q$ ,那么 $a_n, b_n$ 完全由 $b_{n-1}$ 决定, 而我们又有 $b_n&lt;10q$ .由鸽笼原理可知, 在 ${b_1, b_2,\dots, b_{10q}}$ 中必然存在两项相同, 假设这两项是 $b_i,b_j$ , $i&lt;j$ , 那么循环节就是 $b_ib_{i+1}\cdots b_{j-1}$ , 也就是说我们找到了有理数 $\frac{p}{q}$ 的无尽小数表示.</p>
<p>后一个问题更简单些, 并且其实书中已经给出了求法, 这里不过多叙述.</p>
<h1 id="7"><a href="#7" class="headerlink" title="7"></a>7</h1>

显然是无理数. 当然我们还是严谨的证明一下. 如果是有理数, 那么一定有循环节, 易知我们可以在这串数字中得到任意长度连续的 $0$ , 那么循环节只能由 $0$ 组成, 但是又不存在某位之后的所有数字都是 $0$ (必然会出现 $1$), 因此产生矛盾, 也就是说它是无理数.



<h1 id="8"><a href="#8" class="headerlink" title="8"></a>8</h1>
$$
\begin{aligned}
0.2499\cdots=0.25=\frac{1}{4}\\
\begin{aligned}
0.\dot 3\dot7\dot5\cdot1000&=375+0.\dot3\dot7\dot5\\
0.\dot 3\dot7\dot5&=\frac{375}{999}
\end{aligned}\\
4.\dot5\dot1\dot8=4+0.\dot5\dot1\dot8\\
\begin{aligned}
0.\dot5\dot1\dot8\cdot1000&=518+0.\dot5\dot1\dot8\\
0.\dot5\dot1\dot8&=\frac{518}{999}
\end{aligned}\\
4.\dot5\dot1\dot8=4+\frac{518}{999}=\frac{4514}{999}
\end{aligned}
$$



<h1 id="9"><a href="#9" class="headerlink" title="9"></a>9</h1>

与第七题是一样的证法. 易知我们可以从这串数字中得到任意长度连续的 $0$ , 于是循环节只能由 $0$ 组成, 同样也可以得到任意长度的 pl.$1$ , 于是循环节只能由 $1$ 组成, 矛盾, 于是该数为无理数.



<h1 id="10"><a href="#10" class="headerlink" title="10"></a>10</h1><h2 id="1-1"><a href="#1-1" class="headerlink" title="(1)"></a>(1)</h2>

若 $r,s\not=0$ , 那么令 $r=\frac{p}{q}, s=\frac{p'}{q'}$ , 易得 $\sqrt{2}=-\frac{pq'}{qp'}$ 矛盾.



<h2 id="2-1"><a href="#2-1" class="headerlink" title="(2)"></a>(2)</h2>

###### 若 $r,s,t\not=0$ , 那么令  , 那么有 $r^2=2s^2+3t^2+2st\sqrt{6}$ , 等价于 $\sqrt{6} = \frac{r^2-2s^2-3t^2}{2st}$ 显然是有理数, 矛盾.



<h1 id="11"><a href="#11" class="headerlink" title="11"></a>11</h1>

为了方便讨论, 我们将左边的展开式的项分类, 有几个 $a$ 相乘的项就称为几次项. 当它们都大于 $0$ 时, 展开左式得
$$
1+a_1+a_2+\cdots+a_n+a_1a_2+\cdots>1+a_1+a_2+\cdots+a_n
$$
当它们都小于 $0$ 时, 分两种情况讨论.

当 $\sum_{i=1}^n\leqslant-1$ 时. $(1+a_1)(1+a_2)\cdots(1+a_n)>0\geqslant1+a_1+a_2+\cdots+a_n$ .

当 $\sum_{i=1}^n>-1$ 时, 展开左式得 $1+a_1+a_2+\cdots+a_n+S$ , $S$ 是余下的项的和. 现证明 $S>0$ . 显然 $2k$ 次的项的和 (正数) 要大于 $2k+1$ 次项的和 (负数) 的绝对值. 比如
$$
\sum_{i<j} a_ia_j=\sum_{i<j}a_ia_j\cdot1<-\sum_{i<j}a_ia_j\sum_{k\not=i,j}a_k
$$
就是二次项和大于三次项和的绝对值. 也就是说 $S>0$ . 因此原式成立.

有点二项式展开的味道.



<h1 id="12"><a href="#12" class="headerlink" title="12"></a>12</h1>

本题 $(1),(2)$ 题可以合并, 即条件变成 $a_1,a_2,\dots a_n$ 符号相同, 且 $\left|\sum_{i=1}^na_i\right|<1$ . 由 11 题知后半部分成立, 因此我们只需证
$$
\frac{1}{1-\sum_{k=1}^na_k}>\prod_{k=1}^n(1+a_k)
$$
令 $a_{n+1} = 1-\sum_{k=1}^na_k$ , 那么由算数-几何平均不等式得
$$
\begin{aligned}
a_{n+1}\prod_{k=1}^n(1+a_k)&\leqslant1\\
\prod_{k=1}^n(1+a_k)&\leqslant\frac{1}{a_{n+1}}\\
\prod_{k=1}^n(1+a_k)&\leqslant\frac{1}{1-\sum_{k=1}^na_k}
\end{aligned}
$$
显然 $1+a_k>1>a_{n+1}$ , 等号无法取得. 因此
$$
\prod_{k=1}^n(1+a_k)<\frac{1}{1-\sum_{k=1}^na_k}
$$
得证.



<h1 id="13"><a href="#13" class="headerlink" title="13"></a>13</h1>

用书中方法即可证明.



<h1 id="14"><a href="#14" class="headerlink" title="14"></a>14</h1>

分别假设 $a\leqslant b, a>b$ 即可得. 几何意义即是, 在数轴上找到 $a,b$ 的中点, 然后向正半轴大的那个值靠拢.



<h1 id="15"><a href="#15" class="headerlink" title="15"></a>15</h1>

先证明两个的情况.

假设 $\frac{a_2}{b_2}\geqslant\frac{a_1}{b_1}$ , $b_1=kb_2$ , 那么就有$\frac{a_2}{b_2}\geqslant\frac{a_1}{kb_2}$ , 即 $ka_2\geqslant a_1$ , 因此有 $\frac{a_1+a_2}{b_1+b_2}\leqslant\frac{(1+k)a_2}{(1+k)b_2}=\frac{a_2}{b_2}$ , 同理得 $\frac{a_1+a_2}{b_1+b_2}\geqslant\frac{a_1}{b_1}$ .由于对于两项成立, 那么自然对于 $n$ 项成立.



<h1 id="16"><a href="#16" class="headerlink" title="16"></a>16</h1>

题目已经提示了.

当 $n=2$ 时显然成立. 假设当 $n=k$ 时成立, 那么有
$$
(1+x)^k>1+kx
$$
因此
$$
(1+x)^{k+1}>(1+kx)(1+x)=1+(k+1)x+kx^2>1+(k+1)x
$$
因此对于 $n=k+1$ 也成立, 证毕.



<h1 id="17"><a href="#17" class="headerlink" title="17"></a>17</h1>

即证
$$
\begin{aligned}
x^m(y^n-x^n)+y^m(x^n-y^n)&\leqslant0\\
(x^m-y^m)(y^n-x^n)&\leqslant0
\end{aligned}
$$
显然成立 (异号) .



<hr>
<h1 id="题外话"><a href="#题外话" class="headerlink" title="题外话"></a>题外话</h1><p>终于搞定累死我了…</p>
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